Question: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 4} = \dfrac{2x + 8}{x - 4}$
Explanation: Multiply both sides by $x - 4$ $ \dfrac{x^2}{x - 4} (x - 4) = \dfrac{2x + 8}{x - 4} (x - 4)$ $ x^2 = 2x + 8$ Subtract $2x + 8$ from both sides: $ x^2 - (2x + 8) = 2x + 8 - (2x + 8)$ $ x^2 - 2x - 8 = 0$ Factor the expression: $ (x - 4)(x + 2) = 0$ Therefore $x = 4$ or $x = -2$ At $x = 4$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 4$, it is an extraneous solution.